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#!/bin/bash
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: '
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方法一
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LeapYear=`date +%Y`
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if [ $[ $LeapYear % 400 ] -eq 0 ] ; then
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echo "闰年"
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elif [ $[ $LeapYear % 4 ] -eq 0 ] ; then
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if [ $[ $LeapYear % 100 ] -ne 0 ] ; then
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echo "闰年"
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else
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echo "不是闰年"
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fi
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else
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echo "不是闰年"
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fi
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'
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: '
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# 方法二，利用布尔运算实现
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LeapYear=`date +%Y`
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if (( ( "$LeapYear" % 400 ) == "0" )) || (( ( "$LeapYear" % 4 == 0 ) && ( "$LeapYear" % 100 != "0" ) )) ; then
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echo "闰年"
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else
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echo "不是闰年"
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fi
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# 这相当于let语句。如果你尝试类似$ [$ year％400]的东西，你会在这里使用方括号卡住，因为在这里，方括号本身并不代表实际的命令。
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'
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# 利用read命令进行交互式判断实例
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echo "请输入您想要判断是否为闰年的日期(四个数字),然后按enter："
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read LeapYear
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if (( ("$LeapYear" % 400 ) == 0 )) || (( ("$LeapYear" % 4 == 0 ) && ( "$LeapYear" % 100 != 0 ) )); then
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echo "闰年"
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else
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echo "不是闰年"
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fi
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